How to check for differentiability of function of several variables by using definition of differentiability? Show that $F(x,y)= \sqrt{x^2+y^2}$ is not differentiable at $(0,0)$.
In order to do so, i have to arrive at contradiction in the definition. That is, 
$df=f(h,k)-f(0,0)=Ah+Bk+h¥+k₩$ where $A$ is first order partial derivative with respect to $x$ and $B$ is with respect to $y$. And $¥,₩$ are functions in $(h,k)$ which tend to zero as $(h,k)$ tend to zero.
Now, i arrived at $1=\sin\theta+\cos\theta$.
But how should i contradict this last statement? (Coz this is valid for $\theta=0$, 90°)
Also, how to conclude that a function is differentiable through this definition?
 A: Let’s look at what this definition means. For a function $f$ to be differentiable at the point $(x_0,y_0)$, we can find two real numbers $A$ and $B$ such that the change in $f$’s value near that point, $\Delta f=f(x_0+h,y_0+k)-f(x_0,y_0)$ is approximated by $Ah+Bk$, with the error in this approximation vanishing “faster” than the displacement $(h,k)$.  
Let’s apply this to $F:(x,y)\mapsto\sqrt{x^2+y^2}$. The error in any linear approximation to $F$ at the origin is $$\phi(h,k)=\sqrt{h^2+k^2}-Ah-Bk.$$ Along the $x$-axis this error is: $$\phi(h,0)=\sqrt{h^2}-Ah=|h|-Ah.$$ For $h>0$, this is $(1-A)h$. By your definition, we must have $(1-A)\to0$ as $h\to 0$, but $A$ is a constant, so the only way this can happen is if $A=1$. On the other hand, for $h<0$, $$\phi(h,0)=-h-Ah=-(1+A)h=-2h$$ since we’ve already determined that $A$ must be $1$ to make the error along the positive $x$-axis vanish fast enough. Obviously, $-2$ does not tend to zero. We can’t even find a value of $A$ that will produce a small enough error along the $x$-axis only, so there’s no hope of finding $A$ and $B$ that will work everywhere around the origin.
A: Hint: If $F$ were differentiable at $(0,0)$ then $\partial F/ \partial x(0,0)$ would exist. But $F(x,0) = |x|.$
